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Leon Etienne 2022-05-13 11:58:34 +02:00
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@ -90,12 +90,11 @@ First up, we have to establish what requirements this transformation must fulfil
Let's be honest, I'm not a cryptographer, I have no idea how collision resistant this is. Let's be honest, I'm not a cryptographer, I have no idea how collision resistant this is.
This means, it has to be considered *insecure*! This means, it has to be considered *insecure*!
I have tried a few passwords brute-forcibly, experimentally (about 1mil) and have not been able to produce a collision. I have tried a few passwords brute-forcibly, experimentally (about 1mil) and have not been able to produce a collision.
Obviously there have to be collisions, since *|P|, len\(p\) ∈ ℵ ≫ |C|*. Obviously there have to be collisions, since *|P|, len\(p\) ∈ ℵ ≫ |C|, len(c)*.
How does it work? Basically, what happens is your password gets recoded to binary. It is then split into blocks of How does it work? Basically, what happens is your password gets recoded to binary. It is then split into blocks of
size KEY_SIZE, they are &xoplus; together, and this single block is then encrypted with itself as a key. size KEY_SIZE, and they are combined using *c<sub>i+1</sub> = c<sub>i</sub> &xoplus; E(c=block<sub>i</sub>, k=block<sub>i</sub>)*. *c<sub>0</sub>* is a static initialization vector. The final *c* is they key corresponding to a password.
The end result is the key corresponding to your password.
This is a one-way operation. Since the key used for this operation is the cleartext itself, you cannot undo it without already This is a one-way operation. Since the key used for this operation is the cleartext itself, you cannot undo it without already
knowing the password(=cleartext) to begin with. *You could make a hashfunction out of this.* knowing the password(=cleartext) to begin with. *You could make a hashfunction out of this.*